Differentiate the function with respect to $x$: $\frac{\cos^{-1}(\frac{x}{2})}{\sqrt{2x+7}}$,where $-2 < x < 2$.

Let $y = \frac{\cos^{-1}(\frac{x}{2})}{\sqrt{2x+7}}$.
Using the quotient rule $\frac{d}{dx}(\frac{u}{v}) = \frac{v u' - u v'}{v^2}$,we have:
$\frac{dy}{dx} = \frac{\sqrt{2x+7} \cdot \frac{d}{dx}(\cos^{-1}(\frac{x}{2})) - \cos^{-1}(\frac{x}{2}) \cdot \frac{d}{dx}(\sqrt{2x+7})}{2x+7}$
Since $\frac{d}{dx}(\cos^{-1}(\frac{x}{2})) = \frac{-1}{\sqrt{1-(\frac{x}{2})^2}} \cdot \frac{1}{2} = \frac{-1}{\sqrt{4-x^2}}$ and $\frac{d}{dx}(\sqrt{2x+7}) = \frac{1}{2\sqrt{2x+7}} \cdot 2 = \frac{1}{\sqrt{2x+7}}$,we get:
$\frac{dy}{dx} = \frac{\sqrt{2x+7} \cdot (\frac{-1}{\sqrt{4-x^2}}) - \cos^{-1}(\frac{x}{2}) \cdot (\frac{1}{\sqrt{2x+7}})}{2x+7}$
$\frac{dy}{dx} = -\left[ \frac{1}{\sqrt{4-x^2}\sqrt{2x+7}} + \frac{\cos^{-1}(\frac{x}{2})}{(2x+7)^{3/2}} \right]$